<span>So that the ratio is defined:
* The </span><span><span>denominator</span> can not be zero
* Being an integer index pair, the filing must be greater or equal to zero.
It is concluded that:</span>
x ≠ 0 ∧ 2x ≥ 0
x ≥0
.:. x > 0
R/ alternative d) x>0
<u><em>Jeizon1L :)</em></u>
Answer:
Choice A, x=1
Step-by-step explanation:
So, to start, there's two routes that you can take there. I am going to show one of these routes.
Within the image, you see two triangles, you are able to tell that these triangles share two of the same sides, because of congruency lines on each of them.
The first group of congruent sides have 1 line going through them, the second group has 2 lines going through them. Let's go with the first group.
With the first group, we are using the lines CB and FE, where the measurements on each of them are 3x+1 and x+3 respectively.
Since they are congruent, they are equal and we can just set them equal to each other.
Starting With:
3x+1 = x+3
Step 1) Cancel out the +1
3x+1 -1 = x+3 -1
3x = x+2
Step 2) Cancel out the x
3x -x = x+2 -x
2x = 2
Step 3) Divide by 2 get x by itself
2x/2 = 2/2
x=1
All done!
Y² + 8y - 33 :
Break the expression into groups for formula ax²+ bx+c :
= (y²- 3y) + (11y - 33 )
Factor y from y² - 3y => y (y - 3)
Factor out 11 from 11 y - 33 => 11 (y - 3)
= y ( y- 3 ) + 11 ( y - 3 )
Factor out common term (y - 3 ) :
= ( y - 3 ) ( y + 11 )
hope this helps!
Answer:
5x+8
Step-by-step explanation:
2+5+ 3x + 1 + 2x
Combine like terms
3x+2x + 2+5+1
5x +8
Answer:
See below.
Step-by-step explanation:
You fill the 5 gallon up to the top then you pour 3 gallons into the 3 gallon jug, leaving 2 gallons in the bigger jug.
You throw the contents of the 3 gallon jug away then transfer the 2 gallons from the bigger jug into the 3 gallon jug. So you have 2 gallons in the 3 gallon jug.
You then fill the 5 gallon jug and pour 1 gallon into the 3 gallon jug, so you are left with 4 gallons of water in the 5 gallon jug.